If a normal chord of a parabola `y^2 = 4ax` subtends a right angle at the vertex, then it is inclined at angle `theta` with the axis such that `tan^2theta`

Options

A. `1//2`
B. `3//4`
C. 2
D. 4

Answer

To solve the problem, we need to find the value of \( \tan^2 \theta \) for a normal chord of the parabola \( y^2 = 4ax \) that subtends a right angle at the vertex.

### Step-by-Step Solution:

1. **Understanding the Parabola**: The equation of the parabola is given as \( y^2 = 4ax \). The vertex of this parabola is at the origin (0,0).

2. **Normal to the Parabola**: The normal to the parabola at a point \( P(t) \) where \( P(t) = (at^2, 2at) \) has the equation:
\[ y = mx - 2am - at^3 \]
where \( m \) is the slope of the normal.

3. **Finding the Points on the Normal Chord**: Let the points on the normal chord be \( P(t_1) \) and \( P(t_2) \) corresponding to parameters \( t_1 \) and \( t_2 \). The coordinates of these points are:
\[ P(t_1) = (at_1^2, 2at_1) \quad \text{and} \quad P(t_2) = (at_2^2, 2at_2) \]

4. **Condition for Right Angle**: The normal chord subtends a right angle at the vertex, which means the slopes of the lines joining the vertex to these points must satisfy:
\[ m_1 \cdot m_2 = -1 \]
where \( m_1 \) and \( m_2 \) are the slopes of the lines from the origin to \( P(t_1) \) and \( P(t_2) \):
\[ m_1 = \frac{2at_1}{at_1^2} = \frac{2}{t_1}, \quad m_2 = \frac{2at_2}{at_2^2} = \frac{2}{t_2} \]
Thus, we have:
\[ \frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \implies \frac{4}{t_1 t_2} = -1 \implies t_1 t_2 = -4 \]

5. **Finding the Angle \( \theta \)**: The angle \( \theta \) that the normal chord makes with the x-axis can be found using the slope of the normal. The slope of the normal is given by:
\[ m = -\frac{1}{t} \]
Therefore, the angle \( \theta \) can be expressed as:
\[ \tan \theta = -\frac{1}{t} \]

6. **Finding \( \tan^2 \theta \)**: Since we need \( \tan^2 \theta \):
\[ \tan^2 \theta = \left(-\frac{1}{t}\right)^2 = \frac{1}{t^2} \]

7. **Using the Product of Roots**: From the earlier step, we know that \( t_1 t_2 = -4 \). Thus, we can express \( t_1^2 + t_2^2 \) in terms of \( t_1 t_2 \):
\[ t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1 t_2 \]
However, we can directly use \( t_1 t_2 = -4 \) to find \( \tan^2 \theta \).

8. **Final Result**: Since \( t_1 t_2 = -4 \), we can conclude:
\[ \tan^2 \theta = \frac{1}{t^2} = \frac{1}{-4} = \frac{1}{4} \]

Thus, the final answer is:
\[ \tan^2 \theta = 4 \]